{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Times" 1 12 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Courier" 1 14 0 0 0 1 2 1 2 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 0" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 11 255 255 255 1 2 1 2 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 2" -1 257 1 {CSTYLE "" -1 -1 "Courier" 1 10 255 255 0 1 2 1 2 0 0 0 0 0 0 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 123 "This worksheet shows how \+ to draw geodesics on the cone as well as how to calculate the arclengt h of a geodesic. For a=1 in " }{XPPEDIT 18 0 "z=a*sqrt(x^2+y^2" "6#/% \"zG*&%\"aG\"\"\"-%%sqrtG6#,&*$%\"xG\"\"#F'*$%\"yGF.F'F'" }{TEXT -1 203 ", determine c and d for the geodesic connecting (1,0,1) and (0,1, 1) and compare the arclength of this geodesic between these points to \+ the arclength of the part of the parallel circle joining the points." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "with(plots):with(linalg):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "EFG := proc(X)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 " local E,F,G,Xu,Xv;" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 116 " Xu :=[diff(X[1],u),diff(X[2],u),diff(X[ 3],u)];\n Xv := [diff(X[1],v),diff(X[2],v),diff(X[3],v)];" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " E := dotprod(Xu,Xu);" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " F := dotprod(Xu,Xv);" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 " G := dotprod(Xv,Xv);" } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 " simplify([E,F,G]);" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 " end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "The geodesic equations are obtained by" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "geoeq:=proc(X)" }}{PARA 0 "> " 0 " " {MPLTEXT 1 0 28 " local M,eq1,eq2;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 " M:=EFG(X);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 " eq1:=diff(u(t),t$2)+subs(\{u=u(t),v=v(t)\},diff(M[ 1],u)/(2*M[1]))*diff(u(t),t)^2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 89 " \+ +subs(\{u=u(t),v=v(t)\},diff(M[1],v)/(M[1]))*diff( u(t),t)*diff(v(t),t)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 " \+ - subs(\{u=u(t),v=v(t)\},diff(M[3],u)/(2*M[1]))*diff(v(t),t )^2=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 " " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 90 " eq2:=diff(v(t),t$2)-subs(\{u=u(t) ,v=v(t)\},diff(M[1],v)/(2*M[3]))*diff(u(t),t)^2" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 89 " +subs(\{u=u(t),v=v(t)\},diff(M[3 ],u)/(M[3]))*diff(u(t),t)*diff(v(t),t)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 " + subs(\{u=u(t),v=v(t)\},diff(M[3],v)/(2 *M[3]))*diff(v(t),t)^2=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 " \+ eq1,eq2;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 " end: \+ " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 488 "The following procedure plot s a geodesic on a surface. The inputs are: X a parametrization in u an d v, ustart......vend the range of the parameters, u0,v0 the starting \+ point of the geodesic (i.e. initial condition 1), Du0,Dv0 the starting speed of the geodesic (i.e. initial condition 2), n=[a,b,c] where a,b tell what the parameter of the geodesic itself runs from and c denote s how smooth a picture you want, gr=[d,e] where d is the number of gri d lines for u and e is the number for v." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "plotgeo:=proc(X,ustart,uend,vstart,vend,u0,v0,Du0,Dv0 ,T,N,gr,theta,phi)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " local sys, desys,u1,v1,listp,geo,plotX;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 " \+ sys:=geoeq(X);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 75 " desys:=dsolve( \{sys,u(0)=u0,v(0)=v0,D(u)(0)=Du0,D(v)(0)=Dv0\},\{u(t),v(t)\}," }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 " type=numeric, output=listprocedu re);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " u1:=subs(desys,u(t)); v1 :=subs(desys,v(t));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 92 " geo:=spac ecurve(subs(u='u1'(t),v='v1'(t),X),t=0..T, color=black,thickness=2,num points=N):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 80 " plotX:=plot3d(X,u= ustart..uend,v=vstart..vend,grid=[gr[1],gr[2]],shading=XY):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 84 " display(\{geo,plotX\},style=wireframe, scaling=constrained,orientation=[theta,phi]);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "end:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Here is t he cone parametrization for a=1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "CO:=[u*cos(v),u*sin(v),u];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plotgeo(CO,0,1.3,0,2*Pi,1,0,-1,1,1.3,50,[8,30],83, 78);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "The requirement for a co ne geodesic (derived from the fact that the cone is u-Clairaut) is \n \+ " }{XPPEDIT 18 0 "u=C*sec(v/sqrt(1+a^2)+D)" "6#/%\"uG*&%\"CG\" \"\"-%$secG6#,&*&%\"vGF'-%%sqrtG6#,&F'F'*$%\"aG\"\"#F'!\"\"F'%\"DGF'F' " }{TEXT -1 60 ".\nSo, for the geodesic from (1,0,1) to (0,1,1), an an gle of " }{XPPEDIT 18 0 "Pi/2" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 105 " along the z=1 circle, and a=1, we can find D by the following. A t the start and finish, u=1 and v=0 and " }{XPPEDIT 18 0 "Pi/2" "6#*&% #PiG\"\"\"\"\"#!\"\"" }{TEXT -1 62 " respectively. Plugging into the g eodesic formula above gives " }{XPPEDIT 18 0 "1=C*sec(D)" "6#/\"\"\"*& %\"CGF$-%$secG6#%\"DGF$" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "1=C*sec(P i/(2*sqrt(2))+D)" "6#/\"\"\"*&%\"CGF$-%$secG6#,&*&%#PiGF$*&\"\"#F$-%%s qrtG6#F.F$!\"\"F$%\"DGF$F$" }{TEXT -1 30 ". These can be solved to giv e " }{XPPEDIT 18 0 "C=cos(D)" "6#/%\"CG-%$cosG6#%\"DG" }{TEXT -1 5 " a nd " }{XPPEDIT 18 0 "D=arctan((cos(Pi/(2*sqrt(2))-1)/sin(Pi/(2*sqrt(2) ))" "6#/%\"DG-%'arctanG6#*&-%$cosG6#,&*&%#PiG\"\"\"*&\"\"#F/-%%sqrtG6# F1F/!\"\"F/F/F5F/-%$sinG6#*&F.F/*&F1F/-F36#F1F/F5F5" }{TEXT -1 1 "." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "dd:=evalf(arctan((cos(Pi/( 2*sqrt(2)))-1)/sin(Pi/(2*sqrt(2)))));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "cc:=cos(dd);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "So now we can plot the geodesic by replacing u in the parametrization of the cone by the formula above with the C and D just determined." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "geo:=spacecurve([cc*sec(v/ sqrt(2)+dd)*cos(v),cc*sec(v/sqrt(2)+dd)*sin(v)," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 61 " cc*sec(v/sqrt(2)+dd)],v=0..Pi/2,color=black,thick ness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "cone:=plot3d([u *cos(v),u*sin(v),u],u=0..1.3,v=0..2*Pi):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "circ:=spacecurve([cos(v),sin(v),1],v=0..Pi/2,color=re d,thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "display( \{geo,cone,circ\},style=wireframe,scaling=constrained,orientation=[27, 84]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 142 "Finally, we can compute the arclength of the geodesic from (1,0,1) to (0,1,1) and compare it \+ with the arclength of the circle parametrized by " }{XPPEDIT 18 0 "[co s(v), sin(v), 1];" "6#7%-%$cosG6#%\"vG-%$sinG6#F'\"\"\"" }{TEXT -1 121 " joining the two points. Of course, the circle has radius 1, so t he arclength along it from (1,0,1) to (0,1,1) is simply " }{XPPEDIT 18 0 "Pi/2=1.570796327" "6#/*&%#PiG\"\"\"\"\"#!\"\"$\"+Fjzq:!\"*" } {TEXT -1 86 ", since the angle formed by joining the center of the cir cle (0,0,1) to the points is " }{XPPEDIT 18 0 "Pi/2;" "6#*&%#PiG\"\"\" \"\"#!\"\"" }{TEXT -1 29 ". Here is the geodesic again." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "alpha:=[cc/cos(v/sqrt(2)+dd)*cos(v) ,cc/cos(v/sqrt(2)+dd)*sin(v)," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 " \+ cc/cos(v/sqrt(2)+dd)];" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 68 "We take its velocity vector and then compute the a rclength integral." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "alpha p:=diff(alpha,v);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "evalf( Int(sqrt((alphap[1]^2+alphap[2]^2+alphap[3]^2)),v=0..Pi/2));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "So the geodesic has shorter length than the circle." }}}}{MARK "0 0 2" 168 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }